Exercise 1209Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Kernel of a differentiation-like map

#kernel#differentiation#polynomials

Statement

Let $f:\mathbb{R}_3[X]\to\mathbb{R}_3[X]$ be defined by $f(P)=P'$ . Compute $\ker f$ and $\operatorname{Im} f$ .

Hint

$P'=0$ iff $P$ is constant. For the image, every polynomial of degree $\le2$ is the derivative of some polynomial of degree $\le3$ .

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Solution

Kernel. $P'=0$ iff $P$ is constant, so $\ker f=\mathbb{R}$ (the space of constant polynomials). Dimension $1$ .

Image. For any $Q\in\mathbb{R}_2[X]$ , let $P$ be a primitive of $Q$ ; then $P\in\mathbb{R}_3[X]$ and $P'=Q$ . So $\operatorname{Im} f=\mathbb{R}_2[X]$ . Dimension $3$ .

Rank-nullity check. $1+3=4=\dim\mathbb{R}_3[X]$ . Correct.

Conclusion. $\ker f=\mathbb{R}$ and $\operatorname{Im} f=\mathbb{R}_2[X]$ .

Kernel of a differentiation-like map

Statement

Solution

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